a/ => x(x² - 9) = 0 

=> x(x - 3)(x + 3) = 0 

=> x = 0 

hoặc x - 3 = 0 => x = 3

hoặc x + 3 = 0 => x = -3

Vậy x = 0 ; x = 3 ;x = -3

b/ => x² - 6x + x - 6 = 0

=> x(x - 6) + (x - 6) = 0 

=> (x + 1)(x - 6) = 0 

=> x + 1 = 0 => x = -1

hoặc x - 6 = 0 => x = 6

Vậy x = -1 ; x = 6

a)

x(x^2-9)=0

x(x^2-3^2)=0

x(x-3)(x+3)

b) x^2-6x+x-6=0

x(x-6)+(x-6)=0

(x-6)(x+1)=0

1.\(5\left(x^2-9y^2-6y-1\right)=5.\left[x^2-\left(9y^2+6y+1\right)\right]=5\left[x^2-\left(3y+1\right)^2\right]=5\left(x+3y+1\right)\left(x-3y-1\right)\)

2.kiểm tra lại đề nha bạn

3.\(4x^2+10x-2x-5=2x\left(2x-1\right)+5\left(2x-1\right)=\left(2x-1\right)\left(2x+5\right)\)

\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)

\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)

\(=\left(a-b\right)a\left(b-c\right)+\left(b-c\right)c\left(b-a\right)\)

\(=\left(a-b\right)a\left(b-c\right)-\left(b-c\right)c\left(a-b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

mình làm vội, có chỗ nào sai bạn thông cảm nha